Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<long long> v[100005], vt[100005];
long long c[100005];
long long lev[100005], pa[100005][20];
vector<long long> vv[100005][20];
vector<long long> merg(vector<long long> v1, vector<long long> v2,
                       long long a) {
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  vector<long long> tt;
  long long id1 = 0, id2 = 0, cc = 0;
  while (id1 < v1.size() || id2 < v2.size()) {
    if (id1 < v1.size() && id2 < v2.size()) {
      if (v1[id1] < v2[id2]) {
        tt.push_back(v1[id1]);
        id1++;
      } else {
        tt.push_back(v2[id2]);
        id2++;
      }
    } else if (id1 < v1.size()) {
      tt.push_back(v1[id1]);
      id1++;
    } else {
      tt.push_back(v2[id2]);
      id2++;
    }
    cc++;
    if (cc == a) break;
  }
  return tt;
}
void dfs(long long x, long long p, long long le) {
  lev[x] = le;
  for (long long i = 1; i <= 19; i++) {
    if (pa[x][i - 1] != -1) {
      pa[x][i] = pa[pa[x][i - 1]][i - 1];
      vv[x][i] = merg(vv[x][i - 1], vv[pa[x][i - 1]][i - 1], 10);
    }
  }
  for (long long i = 0; i < v[x].size(); i++) {
    if (v[x][i] != p) {
      pa[v[x][i]][0] = x;
      if (vt[v[x][i]].size() != 0) {
        for (long long j = 0;
             j < min((long long)vt[v[x][i]].size(), (long long)10); j++)
          vv[v[x][i]][0].push_back(vt[v[x][i]][j]);
      }
      dfs(v[x][i], x, le + 1);
    }
  }
}
vector<long long> lca(long long u, long long v, long long a) {
  if (lev[u] > lev[v]) swap(u, v);
  long long d = lev[v] - lev[u];
  vector<long long> v1;
  for (long long i = 19; i >= 0; i--) {
    if ((d & (1 << i))) {
      v1 = merg(v1, vv[v][i], a);
      v = pa[v][i];
    }
  }
  if (u == v) {
    vector<long long> tt;
    if (vt[v].size() != 0) {
      for (long long i = 0; i < min((long long)vt[v].size(), a); i++)
        tt.push_back(vt[v][i]);
    }
    v1 = merg(v1, tt, a);
    return v1;
  }
  for (long long i = 19; i >= 0; i--) {
    if (pa[v][i] != pa[u][i]) {
      vector<long long> tt;
      tt = merg(vv[u][i], vv[v][i], a);
      u = pa[u][i];
      v = pa[v][i];
      v1 = merg(v1, tt, a);
    }
  }
  v1 = merg(v1, vv[u][0], a);
  v1 = merg(v1, vv[v][0], a);
  vector<long long> tt;
  if (vt[pa[v][0]].size() != 0) {
    for (long long i = 0; i < min((long long)vt[pa[v][0]].size(), a); i++)
      tt.push_back(vt[pa[v][0]][i]);
  }
  v1 = merg(tt, v1, a);
  return v1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, q;
  cin >> n >> m >> q;
  memset(pa, -1, sizeof(pa));
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (long long i = 1; i <= m; i++) {
    cin >> c[i];
    vt[c[i]].push_back(i);
  }
  dfs(1, 1, 0);
  while (q--) {
    long long u, v, a;
    cin >> u >> v >> a;
    vector<long long> re = lca(u, v, a);
    cout << re.size() << " ";
    for (long long i = 0; i < re.size(); i++) cout << re[i] << " ";
    cout << endl;
  }
  return 0;
}